/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A Radio Tuning Circuit. The mini... [FREE SOLUTION] | 91Ó°ÊÓ

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A Radio Tuning Circuit. The minimum capacitance of a variable capacitor in a radio is 4.18 \(\mathrm{pF}\) . (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(L-C\) circuit is \(1600 \times 10^{3} \mathrm{Hz}\) , corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^{3} \mathrm{Hz}\) . What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?

Short Answer

Expert verified
The inductance is needed to tune the radio and the capacitance should accommodate the given frequency range.

Step by step solution

01

Understand the Problem and Given Data

We need to determine the inductance (L) in a radio tuning circuit given the minimum capacitor value and frequency, and then find the maximum capacitance that allows the range of frequencies in the AM band. We have:\( C_{min} = 4.18 \, \text{pF} = 4.18 \times 10^{-12} \, \text{F} \)\( f_{1} = 1600 \times 10^{3} \text{Hz} \)\( f_{2} = 540 \times 10^{3} \text{Hz} \)
02

Use Frequency Formula for LC Circuit (Part a)

The resonant frequency of an LC circuit is given by \( f = \frac{1}{2 \pi \sqrt{LC}} \). We rearrange this formula to solve for L:\[ L = \frac{1}{(2 \pi f)^{2} C} \] Substitute the values for the minimum capacitance and frequency to find L.
03

Calculate Inductance (L)

Using the formula from Step 2, substitute the given frequency and capacitance:\[ L = \frac{1}{(2 \pi \times 1600 \times 10^{3})^{2} \times 4.18 \times 10^{-12}} \] Calculate L to find the inductance.
04

Calculate Maximum Capacitance (Part b)

The inductance L calculated in Part a is the same for the entire band. To find the maximum capacitance, we use the same formula rearranged to solve for C:\[ C_{max} = \frac{1}{(2 \pi f_{2})^{2} L} \] Substitute the given lowest frequency \( f_{2} = 540 \times 10^{3} \) Hz and the previously calculated L.
05

Perform Calculations for C_max

Calculate the maximum capacitance using the values obtained:\[ C_{max} = \frac{1}{(2 \pi \times 540 \times 10^{3})^{2} \times L} \] Provide the final value of the maximum capacitance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

LC Circuit
An LC Circuit, sometimes called a tuned circuit or *Radio Tuning Circuit*, is an essential component in radios. It consists of an inductor, denoted as L, and a capacitor, labeled as C, connected together. These circuits are fundamentally designed to select specific frequencies from the available numerous radio waves.
This selection of frequencies is possible because the LC circuit can resonate at a particular frequency, which depends on the values of inductance and capacitance used.

Here's how it works:
  • The inductor stores energy in a magnetic field, while the capacitor stores energy in an electric field.
  • When placed in a circuit, energy oscillates back and forth between the inductor and the capacitor.
  • This oscillation occurs at a specific frequency, known as the resonant frequency.
Understanding this mechanism helps in manipulating radio frequencies, allowing the tuner to differentiate between different radio stations. It’s essentially the heart of your radio that finds and locks onto radio signals within the airwaves.
Resonant Frequency
The resonant frequency is crucial in the operation of an LC circuit.It is defined as the frequency at which the circuit naturally oscillates.This frequency is determined by the values of the inductor and the capacitor within the circuit.

Mathematically, the resonant frequency, \( f \), is given by the formula:\[ f = \frac{1}{2 \pi \sqrt{LC}} \]
This equation shows how the resonant frequency depends inversely on the square root of the product of inductance L and capacitance C.

Some applications of the resonant frequency include:
  • In radios, this frequency determines which radio station is selected.
  • In wireless communications, it helps in tuning to different channels.
The ability to adjust the resonant frequency by changing the values of L and C is what makes LC circuits indispensable in frequency tuning applications, especially in communications.
Inductance Calculation
Calculating inductance, or L, in an LC circuit is a vital step when working with tuning circuits.In this part of the process, knowing the resonant frequency and the capacitance allows us to find L.

To find the inductance, the resonant frequency formula is rearranged:\[ L = \frac{1}{(2 \pi f)^2 C} \]
Let's apply this with the given parameters:
  • Given: Frequency \( f \) of 1600 kHz (which is \( 1600 \times 10^3 \) Hz)
  • Capacitance \( C \) of 4.18 pF (which is \( 4.18 \times 10^{-12} \) F)
  • We calculate L by substituting these values into the above formula.
This calculation allows us to find the correct inductance needed for the circuit so that it can achieve resonance at the specified frequency.This process is essential when optimizing or designing radio circuits to ensure they work effectively at desired frequencies.
Capacitance Calculation
Finding the maximum capacitance in an LC circuit expands its ability to cover a range of frequencies.Once the inductance L is known, the maximum capacitance \( C_{max} \) can be calculated for the lowest frequency range applicable.

Here's how you make this calculation:\[ C_{max} = \frac{1}{(2 \pi f_{2})^2 L} \]
Where \( f_{2} \) is the lower band limit frequency, in this case, 540 kHz or \( 540 \times 10^3 \) Hz.With the inductance calculated earlier, just place the numbers in the formula to find \( C_{max} \).This step ensures the radio tuning circuit can accommodate the entire broadcast spectrum it is designed for.

Practical uses in:
  • Wideband radios, offering a complete range of broadcast frequencies.
  • Achieving smooth transitions between different frequency bands.
By performing these calculations, radio circuits can be finely tuned to enhance reception quality across available stations.

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Most popular questions from this chapter

A \(35.0-\mathrm{V}\) battery with negligible internal resistance, a 50.0- \(\Omega\) resistor, and a \(1.25-\mathrm{mH}\) inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed. (a) How long after closing the switch will the current through the inductor reach one-half of its maximum value? (b) How long after closing the switch will the energy stored in the inductor reach one-half of its maximum value?

An inductor with an inductance of 2.50 \(\mathrm{H}\) and a resistance of 8.00\(\Omega\) is connected to the terminals of a battery with an emf of 6.00 \(\mathrm{V}\) and negligible intermal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is \(0.500 \mathrm{A} ;(\mathrm{c})\) the current 0.250 \(\mathrm{s}\) after the circuit is closed; (d) the final steady-state current.

An air-filled toroidal solenoid has 300 turns of wire, a mean radius of \(12.0 \mathrm{cm},\) and a cross-sectional area of \(4.00 \mathrm{cm}^{2} .\) If the current is 5.00 \(\mathrm{A}\) , calculate: (a) the magnetic field in the solenoid; (b) the self-inductance of the solenoid; (c) the energy stored in the magnetic field; (d) the energy density in the magnetic field. (e) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

Two coils have mutual inductance \(M=3.25 \times 10^{-4} \mathrm{H}\) . The current \(i_{1}\) in the first coil increases at a uniform rate of 830 \(\mathrm{A} / \mathrm{s}\) . (a) What is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

A \(15.0-\Omega\) resistor and a coil are connected in series with a 6.30-V battery with negligible internal resistance and a closed switch. (a) At 200 ms after the switch is opened the current has decayed to 0.210 A. Calculate the inductance of the coil. (b) Calculate the time constant of the circuit. (c) How long after the switch is closed will the current reach 1.00\(\%\) of its original value?

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